Integrand size = 56, antiderivative size = 17 \[ \int \frac {e^{\frac {100+e^3}{1+2 x+x^2}} \left (-999+e^3 (-10-2 x)-197 x+3 x^2+x^3\right )}{1+3 x+3 x^2+x^3} \, dx=e^{\frac {100+e^3}{(1+x)^2}} (5+x) \]
Time = 0.32 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {100+e^3}{1+2 x+x^2}} \left (-999+e^3 (-10-2 x)-197 x+3 x^2+x^3\right )}{1+3 x+3 x^2+x^3} \, dx=e^{\frac {100+e^3}{(1+x)^2}} (5+x) \]
Integrate[(E^((100 + E^3)/(1 + 2*x + x^2))*(-999 + E^3*(-10 - 2*x) - 197*x + 3*x^2 + x^3))/(1 + 3*x + 3*x^2 + x^3),x]
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{\frac {100+e^3}{x^2+2 x+1}} \left (x^3+3 x^2-197 x+e^3 (-2 x-10)-999\right )}{x^3+3 x^2+3 x+1} \, dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{\frac {100+e^3}{x^2+2 x+1}} \left (x^3+3 x^2-197 x+e^3 (-2 x-10)-999\right )}{(x+1)^3}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{\frac {100+e^3}{x^2+2 x+1}} \left (x^3+3 x^2-\left (197+2 e^3\right ) x-10 e^3-999\right )}{(x+1)^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (e^{\frac {100+e^3}{x^2+2 x+1}}-\frac {2 \left (100+e^3\right ) e^{\frac {100+e^3}{x^2+2 x+1}}}{(x+1)^2}-\frac {8 \left (100+e^3\right ) e^{\frac {100+e^3}{x^2+2 x+1}}}{(x+1)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int e^{\frac {100+e^3}{x^2+2 x+1}}dx-2 \left (100+e^3\right ) \int \frac {e^{\frac {100+e^3}{x^2+2 x+1}}}{(x+1)^2}dx+4 e^{\frac {100+e^3}{x^2+2 x+1}}\) |
Int[(E^((100 + E^3)/(1 + 2*x + x^2))*(-999 + E^3*(-10 - 2*x) - 197*x + 3*x ^2 + x^3))/(1 + 3*x + 3*x^2 + x^3),x]
3.28.6.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Time = 0.46 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94
method | result | size |
risch | \({\mathrm e}^{\frac {{\mathrm e}^{3}+100}{\left (1+x \right )^{2}}} \left (5+x \right )\) | \(16\) |
gosper | \(\left (5+x \right ) {\mathrm e}^{\frac {{\mathrm e}^{3}+100}{x^{2}+2 x +1}}\) | \(21\) |
parallelrisch | \(x \,{\mathrm e}^{\frac {{\mathrm e}^{3}+100}{x^{2}+2 x +1}}+5 \,{\mathrm e}^{\frac {{\mathrm e}^{3}+100}{x^{2}+2 x +1}}\) | \(38\) |
norman | \(\frac {x^{3} {\mathrm e}^{\frac {{\mathrm e}^{3}+100}{x^{2}+2 x +1}}+11 x \,{\mathrm e}^{\frac {{\mathrm e}^{3}+100}{x^{2}+2 x +1}}+7 x^{2} {\mathrm e}^{\frac {{\mathrm e}^{3}+100}{x^{2}+2 x +1}}+5 \,{\mathrm e}^{\frac {{\mathrm e}^{3}+100}{x^{2}+2 x +1}}}{\left (1+x \right )^{2}}\) | \(86\) |
derivativedivides | \(\left (1+x \right ) {\mathrm e}^{\frac {{\mathrm e}^{3}+100}{\left (1+x \right )^{2}}}+i \sqrt {{\mathrm e}^{3}+100}\, \sqrt {\pi }\, \operatorname {erf}\left (\frac {i \sqrt {{\mathrm e}^{3}+100}}{1+x}\right )-\frac {100 i \sqrt {\pi }\, \operatorname {erf}\left (\frac {i \sqrt {{\mathrm e}^{3}+100}}{1+x}\right )}{\sqrt {{\mathrm e}^{3}+100}}+\frac {400 \,{\mathrm e}^{\frac {{\mathrm e}^{3}}{\left (1+x \right )^{2}}+\frac {100}{\left (1+x \right )^{2}}}}{{\mathrm e}^{3}+100}-\frac {i {\mathrm e}^{3} \sqrt {\pi }\, \operatorname {erf}\left (\frac {i \sqrt {{\mathrm e}^{3}+100}}{1+x}\right )}{\sqrt {{\mathrm e}^{3}+100}}+\frac {4 \,{\mathrm e}^{3} {\mathrm e}^{\frac {{\mathrm e}^{3}}{\left (1+x \right )^{2}}+\frac {100}{\left (1+x \right )^{2}}}}{{\mathrm e}^{3}+100}\) | \(152\) |
default | \(\left (1+x \right ) {\mathrm e}^{\frac {{\mathrm e}^{3}+100}{\left (1+x \right )^{2}}}+i \sqrt {{\mathrm e}^{3}+100}\, \sqrt {\pi }\, \operatorname {erf}\left (\frac {i \sqrt {{\mathrm e}^{3}+100}}{1+x}\right )-\frac {100 i \sqrt {\pi }\, \operatorname {erf}\left (\frac {i \sqrt {{\mathrm e}^{3}+100}}{1+x}\right )}{\sqrt {{\mathrm e}^{3}+100}}+\frac {400 \,{\mathrm e}^{\frac {{\mathrm e}^{3}}{\left (1+x \right )^{2}}+\frac {100}{\left (1+x \right )^{2}}}}{{\mathrm e}^{3}+100}-\frac {i {\mathrm e}^{3} \sqrt {\pi }\, \operatorname {erf}\left (\frac {i \sqrt {{\mathrm e}^{3}+100}}{1+x}\right )}{\sqrt {{\mathrm e}^{3}+100}}+\frac {4 \,{\mathrm e}^{3} {\mathrm e}^{\frac {{\mathrm e}^{3}}{\left (1+x \right )^{2}}+\frac {100}{\left (1+x \right )^{2}}}}{{\mathrm e}^{3}+100}\) | \(152\) |
int(((-2*x-10)*exp(3)+x^3+3*x^2-197*x-999)*exp((exp(3)+100)/(x^2+2*x+1))/( x^3+3*x^2+3*x+1),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\frac {100+e^3}{1+2 x+x^2}} \left (-999+e^3 (-10-2 x)-197 x+3 x^2+x^3\right )}{1+3 x+3 x^2+x^3} \, dx={\left (x + 5\right )} e^{\left (\frac {e^{3} + 100}{x^{2} + 2 \, x + 1}\right )} \]
integrate(((-2*x-10)*exp(3)+x^3+3*x^2-197*x-999)*exp((exp(3)+100)/(x^2+2*x +1))/(x^3+3*x^2+3*x+1),x, algorithm=\
Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {100+e^3}{1+2 x+x^2}} \left (-999+e^3 (-10-2 x)-197 x+3 x^2+x^3\right )}{1+3 x+3 x^2+x^3} \, dx=\left (x + 5\right ) e^{\frac {e^{3} + 100}{x^{2} + 2 x + 1}} \]
integrate(((-2*x-10)*exp(3)+x**3+3*x**2-197*x-999)*exp((exp(3)+100)/(x**2+ 2*x+1))/(x**3+3*x**2+3*x+1),x)
\[ \int \frac {e^{\frac {100+e^3}{1+2 x+x^2}} \left (-999+e^3 (-10-2 x)-197 x+3 x^2+x^3\right )}{1+3 x+3 x^2+x^3} \, dx=\int { \frac {{\left (x^{3} + 3 \, x^{2} - 2 \, {\left (x + 5\right )} e^{3} - 197 \, x - 999\right )} e^{\left (\frac {e^{3} + 100}{x^{2} + 2 \, x + 1}\right )}}{x^{3} + 3 \, x^{2} + 3 \, x + 1} \,d x } \]
integrate(((-2*x-10)*exp(3)+x^3+3*x^2-197*x-999)*exp((exp(3)+100)/(x^2+2*x +1))/(x^3+3*x^2+3*x+1),x, algorithm=\
999/2*e^(e^3/(x^2 + 2*x + 1) + 100/(x^2 + 2*x + 1))/(e^3 + 100) + integrat e((x^3 + 3*x^2 - x*(2*e^3 + 197) - 10*e^3)*e^(e^3/(x^2 + 2*x + 1) + 100/(x ^2 + 2*x + 1))/(x^3 + 3*x^2 + 3*x + 1), x)
Leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (15) = 30\).
Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 4.65 \[ \int \frac {e^{\frac {100+e^3}{1+2 x+x^2}} \left (-999+e^3 (-10-2 x)-197 x+3 x^2+x^3\right )}{1+3 x+3 x^2+x^3} \, dx=x e^{\left (-\frac {x^{2} e^{3} + 100 \, x^{2} + 2 \, x e^{3} + 200 \, x}{x^{2} + 2 \, x + 1} + e^{3} + 100\right )} + 5 \, e^{\left (-\frac {x^{2} e^{3} + 100 \, x^{2} + 2 \, x e^{3} + 200 \, x}{x^{2} + 2 \, x + 1} + e^{3} + 100\right )} \]
integrate(((-2*x-10)*exp(3)+x^3+3*x^2-197*x-999)*exp((exp(3)+100)/(x^2+2*x +1))/(x^3+3*x^2+3*x+1),x, algorithm=\
x*e^(-(x^2*e^3 + 100*x^2 + 2*x*e^3 + 200*x)/(x^2 + 2*x + 1) + e^3 + 100) + 5*e^(-(x^2*e^3 + 100*x^2 + 2*x*e^3 + 200*x)/(x^2 + 2*x + 1) + e^3 + 100)
Time = 9.60 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.82 \[ \int \frac {e^{\frac {100+e^3}{1+2 x+x^2}} \left (-999+e^3 (-10-2 x)-197 x+3 x^2+x^3\right )}{1+3 x+3 x^2+x^3} \, dx={\mathrm {e}}^{\frac {100}{x^2+2\,x+1}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^3}{x^2+2\,x+1}}\,\left (x+5\right ) \]